$f(x) = \cos x - 1 + \frac{x^2}{2!}, x \in R$. Then $f(x)$ is

For the function $f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right)$,where $x \in\left[0, \frac{\pi}{2}\right]$,consider the following two statements:
$(I)$ $f$ is increasing in $\left(0, \frac{\pi}{2}\right)$.
$(II)$ $f^{\prime}$ is decreasing in $\left(0, \frac{\pi}{2}\right)$.
Which of the following is correct?

The function $f(x) = \log(1+x) - \frac{2x}{2+x}$ is increasing on

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} -\frac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^x, & x \leq 0 \end{cases}$. Then $f$ is an increasing function in the interval:

Let $f(x) = \int {e^x}(x - 1)(x - 2)dx$. Then $f$ decreases in the interval

Let $f:(-1, \infty) \rightarrow \mathbb{R}$ be defined by $f(0)=1$ and $f(x)=\frac{1}{x} \ln(1+x), x \neq 0$. Then the function $f$